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No, introduction to mathematical analysis. Charles Matthews 21:50, 7 May 2004 (UTC)Reply

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In the angle defined section, phi is first defined as the area of the sector, then later as the definition of the angle. Shouldn't the angle be twice the sector area (2*phi) or am I missing something? TheGrifter80 (talk) 14:19, 18 August 2025 (UTC)Reply

Correction has been made. Image had a phi so another with angle alpha has replaced it. Text uses variable names as in Hardy's text. — Rgdboer (talk) 21:15, 18 August 2025 (UTC)Reply

Understood now. Thanks for the clarification. TheGrifter80 (talk) 09:53, 19 August 2025 (UTC)Reply

The idea that angle can be defined via area has had several proponents of angle as area. Hardy is one of these as his development in this book is as follows:

A feature of A Course of Pure Mathematics is the definition of angle in terms of an integral. The angle is formed by a line of slope m with the horizontal axis (page 317). With 0 < μ < 1, the point ${\displaystyle P(\mu ,m\mu )}$ is on the unit circle when ${\displaystyle \mu ^{2}+(m\mu )^{2}=1}$. Three equivalent equations are used by Hardy in the demonstration:

${\displaystyle \mu ={\frac {1}{\sqrt {1+m^{2}}}},\quad {\sqrt {1-\mu ^{2}}}={\frac {m}{\sqrt {1+m^{2}}}},\quad m={\frac {\sqrt {1-\mu ^{2}}}{\mu }}.}$

As the derivative of an integral is the integrand, and the derivative of a definite integral is the integrand evaluated at the initial end of the interval of integration, Hardy uses

${\displaystyle {\frac {d}{d\mu }}\int _{\mu }^{1}{\sqrt {1-x^{2}}}dx=-{\sqrt {1-\mu ^{2}}}}$.

With A = (1,0), the area of circular sector POA is ${\displaystyle \phi (m)={\frac {1}{2}}m\mu ^{2}+\int _{\mu }^{1}{\sqrt {1-x^{2}}}dx={\frac {1}{2}}\mu {\sqrt {1-\mu ^{2}}}+\int _{\mu }^{1}{\sqrt {1-x^{2}}}dx}$.

${\displaystyle {\frac {d\phi }{d\mu }}={\frac {1}{2}}{\sqrt {1-\mu ^{2}}}+{\frac {1}{2}}\mu (1-\mu ^{2})^{-1/2}(-2\mu )-{\sqrt {1-\mu ^{2}}}=-{\frac {\mu ^{2}}{\sqrt {1-/u^{2}}}}-{\frac {1}{2}}{\sqrt {1-\mu ^{2}}}={\frac {1}{2{\sqrt {1-\mu ^{2}}}}}.}$

${\displaystyle {\frac {d\phi }{dm}}={\frac {d\phi }{d\mu }}{\frac {d\mu }{dm}}={\frac {1}{2{\sqrt {1-\mu ^{2}}}}}{\frac {m}{(1+m^{2})^{3/2}}}={\frac {1}{2(1+m^{2})}}.}$

As the angle POA is defined as twice the area of its sector in the unit circle, Hardy's definition gives the angle value as ${\displaystyle \int _{0}^{m}{\frac {1}{1+x^{2}}}dx}$.

——As a significant contribution to the foundations of mathematics, it is thought due regard for this passage to be included in the book's article. — Rgdboer (talk) 02:24, 23 April 2026 (UTC)Reply